Question

Find a unit vector perpendicular to each of the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ where $\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$.

Answer 1

Given, $\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}$

and $\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$.

Now, $\overrightarrow{a}+\overrightarrow{b}=(3\hat{i}+2\hat{j}+2\hat{k})+(\hat{i}+2\hat{j}-2\hat{k})=4\hat{i}+4\hat{j}+0\hat{k}$

$\overrightarrow{a}-\overrightarrow{b}=(3\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-2\hat{k})=2\hat{i}+0\hat{j}+4\hat{k}$

We know that the unit vector perpendicular to both $\overrightarrow{a}+\overrightarrow{b}$ ane $\overrightarrow{a}-\overrightarrow{b}$ is given by

$\hat{n}=\frac{(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})}{|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})|}$

Here, $(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 4& 0\\ 2& 0& 4\end{array}\right|$

$=\hat{i}(16-0)-\hat{j}(16-0)+\hat{k}(0-8)$

$=16\hat{i}-16\hat{j}-8\hat{k}$

Now, $|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})|=\sqrt{{\left(16\right)}^2+{\left(16\right)}^2+{\left(8\right)}^2}$

$=\sqrt{256+256+64}=\sqrt{576}=24$

Hence, the required unit vector $\hat{n}=\frac{(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})}{|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})|}$

$=\frac{16\hat{i}-16\hat{j}-8\hat{k}}{24}=\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}$